用定积分的换元积分法求上限1⼀2,下限√2⼀2 ,1⼀x눀√1-x눀dx?

2024-11-01 15:28:05
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回答1:

定积分换元法,令x=sint,t∈[π/6,π/4],则dx=costdt,√(1-x²)=√(1-sin²t)=√cos²t=|cost|=cost,t=arcsinx,当x=½时,t=π/6;当x=√2/2时,t=π/4。于是,原式=